Net impulse is $I=\int_{0}^{t} F d t=m \Delta \vec{v}_{q}$

Initial Potential energy $q$ is $U_{i}=\frac{1}{4 \pi \in_{0}} \frac{Q q}{r}$

Final Potential energy of $q$ is $U_{f}=\frac{1}{4 \pi \in_{0}} \frac{Q q}{2 r}$

As the charge $q$ is only under the influence of electrostatic forces, mechanical energy is conserved.

Final Kinetic energy id $K_{f}=U_{i}-U_{f}=\frac{1}{4 \pi \in_{0}} \frac{Q q}{2 r}$

Thus, $v_{q}=\sqrt{\frac{2 K_{f}}{m}} \Rightarrow I=m v_{q}=\sqrt{2 m K_{f}}=\sqrt{\frac{Q q m}{4 \pi \in_{0} r}}$